CS Operating system Summer Midterm I -- July 11, 2017 You have 115 min (10:00am-11:55pm). Good Luck!

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1 Name / ID (please PRINT) Seq#: Seat #: CS Operating system Summer Midterm I -- July 11, 2017 You have 115 min (10:00am-11:55pm). Good Luck! This is a closed book/note examination. But You can use C reference card(s) given to you.. This exam has 6 questions in 12 pages. Please read each question carefully and answer all the questions, which have 100 points in total. Feel free to ask questions if you have any doubts about questions. Partial credit will be given, so do not leave questions blank. You can get 2pt bonus credit if you complete the boldfaced two columns of the grading table below. Please do this after answering the questions in the exam. Thanks! Difficulty level of Question Topic Possible Points this question 1: Easiest 5: Most difficulty Student Expects out of 1 Review questions, short 20 /20 explanations 2 Program and Process, 15 /15 static keyword, cmd line 3 CPU Scheduling 15 /15 Student Received 4 Unix I/O and file 15 /15 operations, fork/exec 5 IPC: pipes -fifo 15 /15 6 Memory 20 /20 Bonus 2 Total

2 1. [20 points] Answer the following questions with short explanations. If needed, you can draw figures to better explain your ideas. a. [2pt] What does van Neumann Architecture mean? b. [2pt] Explain the difference between a System Call (e.g., write) and a Library Function Call (e.g., printf). You can draw figures to better explain your steps (control flow) in carrying out the System Calls or Library Function Calls.. c. [2pt] Describe the actions taken by a kernel to context-switch between processes. 2

3 d. [2pt] List all the possible output sequences for the following program. void main() { if (fork()!= 0) { printf("a"); waitpid(-1, NULL, 0); else { printf("b"); printf("c"); exit(0); possible outputs e. [2pt] Give the process states and show the possible state transitions on a diagram. f. [2pt] In Round Robin Scheduling, the quantum size is an important parameter to decide. Explain what are the advantages/disadvantages of making it too long or too short. 3

4 g. [2pt] Recall the INODE structure which uses some direct pointers, and then single, double, and triple indirect pointers to access data blocks. What is the reasoning behind this design choice, and what are the advantages/disadvantages of having multi-level indirections? h. [2pt] Suppose you execute the following two statements in a program without any error: myfd1 = open("my.dat", O_RDONLY); myfd2 = open("my.dat", O_RDONLY); Show the contents of myfd1 and myfd2, and more importantly the contents of and the relationship between the file descriptor table, the system file table and in-memory inode table. 4

5 i. [2pt] What are the similarities and/or differences between unnamed pipe and named pipe (FIFO)? j. [2pt] Suppose the following code sections are executed without any error. Draw the diagram showing the relationship between the file descriptors, pipe and process (first one is given). Code section int fd[2]; pipe(fd); Diagram after the execution of the code section dup2(fd[0], STDIN_FILENO); dup2(fd[1], STDOUT_FILENO); close(fd[0]); close(fd[1]); 5

6 2. [15 points] Program and Process, command-line arguments, static keyword Consider the following program, which calls a function to merge the given command line arguments (except the program name) as a dynamically created single string. It inserts ':' as the separator between the command line arguments in the new string. Finally, the program prints the returned string and then frees it. #include <stdio.h> #include <stdlib.h> char *merge_command_line_args(int argc, char **argv); int main(int argc, char *argv[]){ char *str; str = merge_command_line_args(argc, argv); printf("returned string is %s\n", str); free(str); return 0; If we run this program as > prog aaaa bbbb ccc It should simply create the following string and print it as aaaa:bbbb:ccc Now you are asked to implement the abovementioned function that dynamically creates a new string by merging the given command line arguments and separating them with ':'. You can use any standard library functions if needed. char *merge_command_line_args(int argc, char **argv) { 6

7 3. [15 Points] CPU Scheduling Consider two processes as in Assignment 2/Quiz 4, where each process has two CPU bursts with one I/O burst in between on a single core CPU. Suppose P1 and P2 have the following life-cycles: P1 has x1=4, y1=1, z1=7 units for the first CPU burst, I/O burst, second CPU burst, respectively. P2 has x2=3, y2=7, z2=2 units for the first CPU burst, I/O burst, second CPU burst, respectively. Both arrives at the same time (in case of ties, pick P1) and there is no other processes in the system. For each of the scheduling algorithms below, create Gantt charts as you did for the Quiz 4. Fill each box with the state of the corresponding process. Use R for Running, W for Waiting for I/O, and r for waiting in ready queue. Calculate the average waiting times in ready queue and CPU utilization for each algorithm. Gantt Charts for SJF (Shortest Job First, non-preemptive) [3pt] a) SJF P1 P2 Gantt Charts for PSJF (Preemptive SJF) [3pt] b) PSJF P1 P2 Gantt Charts for Round Robin (RR with quantum = 3 ) [3pt] b) RR P1 P2 Compute average waiting time in ready queue and CPU utilization for each of three algorithms [6pt] 7

8 4. [15 points] Unix I/O and file operations a. [8 points] Suppose that the file tmpdata.txt contains abcdefghijk. Suppose the following code is executed correctly without generating any errors. 1: int fd; 2: char buf[6] = ; 3: fd = open( tmpdata.txt, O_RDONLY); 4: if (fork( )==0) fork(); 5: read(fd, buf, 2); 6: read(fd, buf+2, 2); 7: printf( %d: %s, (long)getpid(), buf); i). [4pt] Explain if the following two outputs are possible or not and Why/why not? Suppose parent's pid is 7 while child's pid is 8 and the pid of the child's child is 9. 7: a2bc5 8: a2b45 9: abc45 7: a2ef5 8: b2g45 9: cdhi5 ii). [4pt] Exchange the lines 3 and 4! Explain if the above two outputs are possible now (why/not)? b. [7 points] Suppose you are given the inode structure as shown below. Assume that our inode structure has 10 direct pointers, and Block size is 8K bytes and pointers are 4 bytes. Write a function int count_ti(struct inode *myinode, char *word); which counts the number of blocks containing the given string word only in triple indirect blocks. db si di ti Note that triple indirect level might be partially filled with blocks! So if there is no more block or level the corresponding pointer will contain NULL. Please take that into account! Also assume that a data block contains a string text. Thus, if needed, you can use strcmp() or any other functions from the standard libraries (suppose they are included here). 8

9 Space for problem 4 (b) [7pt] int count_di(struct inode *myinode, char *word){ int i, j, k; int count=0; int num_blk = 8 * 1024 / 4; 9

10 5. [15 points] IPC and pipes -fifo, fork/exec Suppose we have three programs aaa, bbb, ccc which are already developed, compiled and ready to execute. The first two, namely aaa and bbb, simply do some random amount of computation and write a character (respectively, 'A' and 'B') to STDOUT_FILENO in a loop that is repeated 1000 times. If we run both programs at the same time, we might see different numbers of As and Bs such as AABBBAAABBBBABB... on the screen. The third program, namely ccc, reads a sequence of characters from STDIN_FILENO and simply counts the number of the same characters that appear consecutively and writes that information to STDOUT_FILENO. For example, if we give the above possible sequence of As and Bs to ccc, it will print A 2, B 3, A 3, B 4, A 1, B 2,... on the screen. Now you are asked to write a program (say xyz.c) which can execute aaa, bbb, and ccc as children and connect them in a way that the output of aaa and bbb will go into a single pipe. Then, ccc will get the mixed sequence of characters from this pipe and do its job. Your program (parent) will wait for all three children to be done, then it will quit! Convince yourself that your program should create the following relation through a pipe! child1(aaa)[1] ----> mypipe ---> [0]child3(ccc) child2(bbb)[1] ----> You can ignore most of the error checking to make your code clear, but you need to close all unnecessary file/pipe descriptors and need to check what fork() returns to detect child/parent proceees and/or what read()/write() returns to detect end of file etc. To execute a given program simply use execl("aaa","aaa",null); etc. /* your implementation of xyz.c */ #include <fcntl.h> #include <stdio.h> #include <unistd.h> #include <sys/stat.h> int main (int argc, char *argv[]) { int mypipe[2]; int child1=1, child2=1, child3=1; /* parent: create mypipe and children processes (3pt) */ /* continue problem 5 in the next page */ 10

11 if(child1 == 0){ /* child 1: set up pipe and exec aaa (3pt) */ execl("aaa","aaa",null); perror("cannot start aaa"); return 1; if(child2 == 0){ /* child 2: set up pipe and exec bbb (3pt) */ execl("bbb","bbb",null); perror("cannot start bbbb"); return 1; if(child3 == 0){ /* child 3: set up pipe and exec ccc (3pt) */ execl("ccc","ccc",null); perror("cannot start ccc"); return 1; /* parent: final things to do (3pt) */ 11

12 6. [20 points] Memory Management/Design, Address translation and TLB performance: A. [10 points] Suppose our machine has a 64 Kbytes of virtual address space and is byteaddressable. The physical memory is 16 Kbytes and the page/frame size is 512 bytes. Give the best answer to each of the following with a short justification. If you need to make additional assumptions to give an answer, state the assumptions that you are making [Note: if the answer is a power of 2, you can leave it in exponential form.] a. [1pt ]How many bits are used for the logical/virtual address? b. [1pt] How many bits are used for the physical address? c. [1pt] How many bits are used for the page offset (d)? d. [1pt] How many bits of virtual address are used for page number (p)? e. [1pt] How many bits of physical address are used for frame number (f)? f. [1pt] How many pages are in a process virtual address space? g. [1pt] How many frames are in the physical memory? h. [1pt] How many bytes would be needed for a page table entry? i. [2pt] How many bytes would be needed for a page table of a process? B. [10 points] Suppose we use the following page table and TLB table entries in the above system for memory address translation, where TLB access time is 8ns and memory access time is 22ns. initial page table Index Valid frame TLB Page # Frame # For the logical/virtual addresses given in binary below, find the corresponding page number and frame number while also determining the time to access the actual data/instruction in the physical memory (suppose we always first check TLB and then check page table if the page number is not in TLB). Also give the corresponding physical memory address. Logical address Page # Access time Frame # Physical Address - bitwise (1pt) (4pt) (1pt) (4pt)